Hatrack River Writers Workshop: Question about Time Dilation

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Posted by redux (Member # 9277) on :

I am trying to write a story where time dilation will come into play and I was wondering if someone could help me figure it out.

For instance, say a Tourist wants to travel 1 AU (astronomical unit) which is roughly 150 million kilometers. I am doing away with acceleration and deceleration.

From an Earth observer, if the Tourist travels at the speed of light (or really really close to it) that would be 150 million km divided by 300,000 km/sec which equals 500 seconds or 8 minutes.

From the Tourist's perspective, hardly any time has passed, correct? This is because time slows down for those travelling at the insanely high speeds.

Here's where I get lost...

What I am trying to figure out is how to go in "reverse." Meaning, how much time passes on Earth if a Tourist feels they have been gone for say 30 days travelling at the speed of light?

Please help!

Posted by Brendan (Member # 6044) on :

Ok, at light speed (and the only way they could go at light speed it to be turned into light - nothing with mass can go at light speed), then the journey would feel instantaneous.

If they are traveling near light speed, then it will depend on how close to light speed they travel. If they are traveling at 0.9c, where c is the speed of light, then 158 days would pass for 30 days relative. At 0.999c, 41 years. At 0.999999c, 41000 years.

Posted by LDWriter2 (Member # 9148) on :

Sorry I can't help you. I know it can be done, I've read stories where it has happened. I think it was Robert Heinlein who did a novel where that was one of the features of the story. It was about a set of twins who could communicate over great distances. One went exploring in space while his brother stayed on the Earth. So the idea has been around for a long time.

Posted by DerekBalsam (Member # 8471) on :

Brendan has it right. Another way to look at it is that from the Earth's perspective, the trip always takes the "expected" time. Put simply: 1,000,000,000 km per hour is 93% the speed of light. If you travel 150,000,000 km at 1,000,000,000 km per hour, then from the Earth's point of view the ride takes 9 minutes (150 000 000 km) / (1 000 000 000 (km / hr)). Just like a 30-mile ride at 60 miles/hour will take 30 minutes.

The Earth's time is not dilated, is all, it's just normal.

(This is a simplification that leaves out lots of assumptions about inertial reference frames and simultaneity).

Posted by redux (Member # 9277) on :

Thanks for the replies.

LDWriter2 - I think that Heinlein story is 'Time for the Stars' featuring telepathic twins.

I'm just a bit fuzzy on the math. So, if my Tourist decides to travel at 90% the speed of light for 1 second according to that calculator roughly 2.3 seconds have passed for an Earth observer. So, if I have a character take a 1 year round-trip journey at 0.9 the speed of light the Earth observer shouldn't expect this character to come back until a little over 2 years and 3 months?

I just don't want to offend any physics majors when writing my story

[This message has been edited by redux (edited October 12, 2010).]

Posted by philocinemas (Member # 8108) on :

I recently posted a link to a site that helps with this. However, I plugged in the numbers and came out with a different outcome - not sure who's right.

http://www.cthreepo.com/lab/math1.shtml

Posted by Brendan (Member # 6044) on :

Oops, forgot to do the square root. Look up wiki under Time Dilation for the formula

Try this

V (in c) Earth Time (days)
0.9 . . . . 68.8 days
0.999 . . 671 days
0.999999 . 58.1 years

Or use Philo's website

[This message has been edited by Brendan (edited October 12, 2010).]

Posted by LDWriter2 (Member # 9148) on :

quote:
Thanks for the replies.
LDWriter2 - I think that Heinlein story is 'Time for the Stars' featuring telepathic twins.

I thought of need be you could go to the book to figure it out, I should have realized there would be a site that had the formula, even for something a large percentage of the population would not be interested in.

Posted by redux (Member # 9277) on :

quote:
I thought of need be you could go to the book to figure it out, I should have realized there would be a site that had the formula, even for something a large percentage of the population would not be interested in.

The random things you can find on the internet never ceases to amaze me

Posted by MartinV (Member # 5512) on :

Ok, we actually did this problem in class so I can discuss details.

When you consider relativistic travel, you should not forget that the space ship does not achieve cruising velocity in a blink of an eye. A huge mass needs time to gather momentum and it uses acceleration to do it. Therefore you should really ask yourself what kind of acceleration do you need to use to travel the distance.

Calculating acceleration is quite different than calculating velocity but it does make sense. If you are out in space in null gravity, you can use acceleration to simulate gravity. Therefore accelerating with 1 g makes most sense to gather speed, then halfway to your destination you use negative acceleration to slow the ship down, again creating perfect Earth-like gravity. If you however want to take time travelling with constant velocity (and therefore null gravity which sounds nice until you think about the effects on human physiology), then you need to use the special relativity equations.

We did the math on the famous twin paradox (though it's not a paradox at all) and came up with a following solution. If you start two stopwatches when the space ship leaves Earth (one stopwatch stays on Earth, the other travels with the spaceship), in the middle of the journey (the point where you need to stop increasing velocity and start decreasing it) they will show different results. The difference is:

tE = c/g *cosh(g*tT/c) or inverted
tT = c/g *arccosh(g*tE/2/c)

tE - Earth time
tT - Traveller's time
g - Earth's gravity (9.80665 m/s/s)
c - light speed (299,792,458 m/s)
cosh - hyperbolic cosine function (Windows calculator has it on scientific view; enable the Hyp square and then press the 'cos' button)
arccosh - inverse hyperbolic cosine function (Windows calculator also has it; enable both Inv and Hyp squares, then click 'cos' button)

Now if you want to know the time at reaching the destination, simply multiply the result with 2. If your space ship goes one way, then immediately turns and returns to Earth the same way, just use 4 instead of 2.

Now, for the distance. Let's say you want to travel the distance from Earth to Alpha Centauri, the closest star. The distance is roughly 4 light years. That means you need to accelerate 2 light years and decelerate the other 2 light years. You get halfway (the point where you need to stop increasing velocity and start decreasing it) in the time given by:

tT = c/g *sqrt[ (x*g/c/c)^2 - 1 ]

So if you input x = 2 ly (1 light year = 9.45e15 m) and the rest as it was above, then you get 1.71 years which is about 20 months. The destination to Alpha Centauri occurs in 40 months. To get back to Earth you need 80 months.

How much time passes for the observer on Earth? Put tT into the above equation and you get tE = 2.91 years, almost twice the time.

You can modify the two formulas. If you want to use higher acceleration, input 2g or 3g instead of 1g. The c and g values are absolute, the rest is simple computation. Just remember to input all numbers in the same units, preferably meters and seconds since both g anc c are defined in those units. If you input kilometers and years or god forbid miles, you will get wrong results.

The beauty of these equations is that you don't need to bother with the relativistic factors (gamma = sqrt[ 1 - v*v/c/c] ) at all. You don't even need to know the velocity at which you will be travelling.

Just for reality's sake: it would make little sense for a distance of 1 AU to be travelled at relativistic speeds. When it comes to interstellar distances, then you can talk FTL.

[This message has been edited by MartinV (edited October 14, 2010).]

Posted by redux (Member # 9277) on :

MartinV - thanks for the detailed explanation!